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Question Number 204300 by depressiveshrek last updated on 11/Feb/24

x^2 log_3 x^2 −(2x^2 +3)log_9 (2x+3)=3log_3 ((x/(2x+3)))

$${x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}^{\mathrm{2}} −\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{9}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 11/Feb/24

x^2 log_3 x^2 −(2x^2 +3)log_9 (2x+3)=3log_3 ((x/(2x+3))) log_9 (2x+3)=((log_3 (2x+3) )/(log_3 9 ))=((log_3 (2x+3) )/2) 2x^2 log_3 x−(2x^2 +3)(((log_3 (2x+3) )/(2 )))=3log_3 ((x/(2x+3))) 2x^2 log_3 x−(2x^2 +3)log_3 (2x+3)=6log_3 ((x/(2x+3))) 4x^2 log_3 x−(2x^2 +3)log_3 (2x+3)=log_3 ((x/(2x+3)))^6 log_3 x^(4x^2 ) −log_3 (2x+3)^(2x^2 +3) =log_3 ((x/(2x+3)))^6 log_3 (x^(4x^2 ) ÷(2x+3)^(2x^2 +3) )=log_3 ((x/(2x+3)))^6 x^(4x^2 ) ÷(2x+3)^(2x^2 +3) =((x/(2x+3)))^6 (x^(4x^2 ) /((2x+3)^(2x^2 +3) ))=((x/(2x+3)))^6

$${x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}^{\mathrm{2}} −\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{9}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{log}_{\mathrm{9}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\frac{\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)\:}{\mathrm{log}_{\mathrm{3}} \mathrm{9}\:}=\frac{\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)\:}{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\left(\frac{\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)\:}{\mathrm{2}\:}\right)=\mathrm{3log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)\:\:\: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{6log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)\:\:\: \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \:\:\: \\ $$$$\mathrm{log}_{\mathrm{3}} {x}^{\mathrm{4}{x}^{\mathrm{2}} } −\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}} =\mathrm{log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \:\: \\ $$$$\mathrm{log}_{\mathrm{3}} \left({x}^{\mathrm{4}{x}^{\mathrm{2}} } \boldsymbol{\div}\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}} \right)=\mathrm{log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \\ $$$${x}^{\mathrm{4}{x}^{\mathrm{2}} } \boldsymbol{\div}\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}} =\left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \\ $$$$\frac{{x}^{\mathrm{4}{x}^{\mathrm{2}} } }{\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}} }=\left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \\ $$

Commented by deleteduser1 last updated on 12/Feb/24

⇒x^(4x^2 −6) =(2x+3)^(2x^2 −3) ⇒(x^2 )^(2x^2 −3) =(2x+3)^(2x^2 −3) ⇒((x^2 /(2x+3)))^(2x^2 −3) =1⇒(x^2 /(2x+3))=1∨2x^2 −3=0 ⇒x=3,x=((√6)/2),

$$\Rightarrow{x}^{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{6}} =\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}} \Rightarrow\left({x}^{\mathrm{2}} \right)^{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}} =\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}} \\ $$$$\Rightarrow\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}} =\mathrm{1}\Rightarrow\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{3}}=\mathrm{1}\vee\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3},{x}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}, \\ $$

Commented by Rasheed.Sindhi last updated on 12/Feb/24

Thanks sir!

$$\mathcal{T}{hanks}\:{sir}! \\ $$

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